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Lie bracket of derivations in space of r-forms

  1. Sep 3, 2014 #1
    In textbook by Kobayashi and Nomizu derivation of rank k in space of all differential forms on a manifold is defined to be operator that is linear, Leibnitz and maps r-forms into r+k-forms. By Leinbitz I mean, of course: [itex]D(\omega \wedge \eta)=(D \omega) \wedge \eta + \omega \wedge (D \eta)[/itex]. Skew-derivation is defined similarly, but rather than Leibnitz property it satisfies: [itex]D(\omega \wedge \eta)=(D \omega) \wedge \eta + (-1)^r \omega \wedge (D \eta)[/itex], where [itex]\omega[/itex] is r-form.
    Then claim is made that if [itex]D[/itex] is derivation of rank k and [itex]D'[/itex] is skew-derivation of rank k' then [itex][D,D'][/itex] is skew-derivation of rank k+k'. However, formula which I'm getting (just by using the properties given) is:
    [tex][D,D'] ( \omega \wedge \eta )= \left\{ ( [D,D'] \omega) \wedge \eta + (-1)^r \omega \wedge ([D,D'] \eta) \right\} +(-1)^r (1-(-1)^k) (D \omega) \wedge (D' \eta)[/tex]
    Term outside of curly brackets vanishes only for k even, so as far as I understand, claim given by K&N holds only then. If anyone has an idea what could have gone wrong (or is it possible that there is an error in K&N), please help. Thanks in advance.
  2. jcsd
  3. Sep 3, 2014 #2
  4. Sep 4, 2014 #3
    This problem led me to following conjecture: There is nothing wrong with my calculations, theorem in K&N is true, but the fact they omitted is that for derivations in differential forms spaces k is always even for Leibnitz derivations (example: Lie derivative) and odd for skew derivations (exterior derivative, interior product). That would solve problem above and nearly identical problem I had with anticommutator of two skew-derivations.

    That conjecture is however based on the fact that I know no counterexamples (but I am not an expert) and it just fits. I have no idea how one could approach proving such thing. Maybe some deeper knowledge about graded algebras is needed?
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